Find the magnitude of F, and the tension in the string A particle of weight 40N is attached to the end of a light extensible string of length 2m. the other end of the string is attached to a vertical wall. the particle is held at a distance of 1.2m from the wall by a horizontal force F.

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We have a particle of weight 40N attached to a string 1.2 m from a wall. The string is pivoted on a vertical wall and there is a force of magnitude F acting upwards on the string at a distance of 2m from the walls.

The particle is at equilibrium if the total torque in the system is zero. This is possible when 1.2 * 40 = 2 * F

=> F = 1.2 * 40 / 2

=> F = 0.6 * 40

=> F = 24 N

The magnitude of the force is 24 N.

The tension in the string is 40 N as the tension in the string added to the downward force exerted by the particle has to be equal to zero which is indicated by the lack of acceleration.

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